# Problem of the Week Problem D and Solution Five Digits

## Problem

A sequence starts out with one $$5$$, followed by two $$6$$s, then three $$7$$s, four $$8$$s, five $$9$$s, six $$5$$s, seven $$6$$s, eight $$7$$s, nine $$8$$s, ten $$9$$s, eleven $$5$$s, twelve $$6$$s, and so on. (You should notice that only the five digits from $$5$$ to $$9$$ are used.)

The first $$29$$ terms of the sequence appear below.

$$5$$, $$6$$, $$6$$, $$7$$, $$7$$, $$7$$, $$8$$, $$8$$, $$8$$, $$8$$, $$9$$, $$9$$, $$9$$, $$9$$, $$9$$, $$5$$, $$5$$, $$5$$, $$5$$, $$5$$, $$5$$, $$6$$, $$6$$, $$6$$, $$6$$, $$6$$, $$6$$, $$6$$, $$7$$, $$…$$

Determine the $$2022^{\text{nd}}$$ digit in the sequence.

Note:
In solving the above problem, it may be helpful to use the fact that the sum of the first $$n$$ positive integers is equal to $$\dfrac{n(n+1)}{2}$$. That is,

$1 + 2 + 3 + … + n = \frac{n(n+1)}{2}$ For example, $$1 + 2 + 3 + 4 + 5 = 15$$, and $$\dfrac{5(6)}{2} = 15$$.

Also, $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$$, and $$\dfrac{8(9)}{2} = 36$$.

## Solution

The first group in the sequence contains one $$5$$. The second group in the sequence contains two $$6$$s. To the end of the second group of digits, there is a total of $$1+2=3$$ digits. The third group in the sequence contains three $$7$$s. To the end of the third group of digits, there is a total of $$1+2+3=6$$ digits. The $$n$$th group in the sequence contains $$n$$ digits. To the end of the $$n$$th group of digits, there is a total of $$1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$$ digits.

How many groups of digits are required for there to be at least $$2022$$ digits in the sequence?

We need to find the value of $$n$$ so that $$1+2+3+\cdots +n \ge 2022$$ and $$1+2+3+\cdots +(n-1) < 2022$$. At this point we will use trial and error. At the end of the solution, a more algebraic approach to finding the value of $$n$$ using the quadratic formula is presented.

Suppose $$n=100$$. Then $$1+2+3+\cdots +100 =\dfrac{100(101)}{2}= 5050>2022$$.

Suppose $$n=50$$. Then $$1+2+3+\cdots +50 =\dfrac{50(51)}{2}= 1275<2022$$.

Suppose $$n=60$$. Then $$1+2+3+\cdots +60 =\dfrac{60(61)}{2}= 1830<2022$$.

Suppose $$n=65$$. Then $$1+2+3+\cdots +65 =\dfrac{65(66)}{2}= 2145>2022$$.

Suppose $$n=63$$. Then $$1+2+3+\cdots +63 =\dfrac{63(64)}{2}= 2016<2022$$.

The $$2022$$nd digit is the sixth number in the next group of digits. That is, the $$2022$$nd digit is a digit in the $$64$$th group of digits.

Now, let’s determine what digit is in the $$64$$th group of digits. Since we cycle through the digits and there are only five digits used, we can determine the digit by examining $$\frac{64}{5} = 12\frac{4}{5}$$. Thus, in the $$64$$th group of digits, the digit used is the $$4$$th digit in the sequence of digits. That is, in the $$64$$th group of digits, the digit used is an $$8$$.

Since the $$2022$$nd digit is in the $$64$$th group of digits, it follows that the $$2022$$nd digit is an $$8$$.

We will finish by showing how we can find the value of $$n$$ algebraically.

We will first find the value of $$n, n>0$$, so that \begin{aligned} \frac{n(n+1)}{2}&=2022\\ n(n+1)&= 4044\\ n^2+n-4044&=0\end{aligned} The quadratic formula can be used to solve for $$n$$. \begin{aligned} n&=\frac{-1\pm\sqrt{1^2-4(1)(-4044)}}{2}\\ &=\frac{-1\pm\sqrt{16\,177}}{2}\end{aligned} Since $$n=\dfrac{-1-\sqrt{16\,177}}{2}<0$$, it is inadmissible.

Then $$n=\dfrac{-1+\sqrt{16\,177}}{2}\approx 63.09$$. But $$n$$ is an integer. So, interpreting our result, when $$n=63$$, the sum $$1+2+3+\cdots+63<2022$$, and when $$n=64$$, the sum $$1+2+3+\cdots+64>2022$$. Thus, the $$2022$$nd digit is in the $$64$$th group of digits.